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Q.

In figure, PA and PB are tangents to the circle with center O such that APB= 50 °  . Write the measure of OAB.  

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a

90 o   

b

50 o  

c

60 o  

d

25 o  

answer is A.

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Detailed Solution

Given that PA and PB are tangents to the circle with center O such that APB= 50 °  .
We have to find OAB.  
The tangent of a circle at any point in a circle is perpendicular to the radius through the point of contact
The required figure geometry is shown below,
Question ImageSince the length of the tangent from an external point to a circle is equal. So, PA=PB   . In triangle PBA,PBA=PAB   . Because angles opposite the equal sides are equal. The sum of the three angles of a triangle is equal to 180 degrees. So, APB+PBA+PAB = 180 ° 50 ° +PAB+PAB = 180 ° 2PAB = 180 ° 50 ° PAB = 65 °   From the theorem, the tangent of a circle at any point in a circle is perpendicular to the radius through the point of contact. So, OAPA   and OAP= 90 °   Thus, PAB+OAB = 90 ° 65 ° +OAB = 90 ° OAB = 90 ° 65 ° OAB = 25 °  
The value of OAB   is 25 °  
Therefore, the correct option is 1.
 
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In figure, PA and PB are tangents to the circle with center O such that ∠APB= 50 °  . Write the measure of ∠OAB.